What is the extreme f(x)=(x^3)/3-4x ?

The extreme f(x)=(x^3)/3-4x is Maximum(-2, 16/3),Minimum(2,-16/3)

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extreme f(x)=(x^3)/3-4x

extreme points

f (x) = \frac{x ^{3}}{3} - 4 x

Maximum (- 2, \frac{16}{3}), Minimum (2, - \frac{16}{3})

Solution steps

f^{'} (x) = x^{2} - 4

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{x ^{3}}{3} - 4 x : \frac{16}{3}

Maximum (- 2, \frac{16}{3})

Plug

x = 2

into

\frac{x ^{3}}{3} - 4 x : - \frac{16}{3}

Minimum (2, - \frac{16}{3})

Maximum (- 2, \frac{16}{3}), Minimum (2, - \frac{16}{3})

f (x) = \frac{x}{6 x ^{2} + 5 x - 1}

f (x) = \frac{3 x}{x + 5}

y = 5 x - 9

f (x) = x^{2} + 5 x + 6

\frac{x ^{2} - 4}{x ^{3}}

What is the extreme f(x)=(x^3)/3-4x ?
The extreme f(x)=(x^3)/3-4x is Maximum(-2, 16/3),Minimum(2,-16/3)